Question

Under the influence of a uniform magnetic field, a charged particle is moving in a circle of the radius $\mathrm{R}$ with constant speed $\mathrm{v}$. The time period of the motion is

• A. depends on $\text{v}$ and not on $\text{R}$
• B. depends on both $\text{R}$ and $\text{v}$
• C. is independent on both $\text{R}$ and $\text{v}$
• D. depends on both $\text{R}$ and not on $\text{v}$

Answer 1

The correct option is C

is independent on both $\text{R}$ and $\text{v}$

The explanation for the correct option:

Step 1: Given data

A charged particle is moving in a circle of the radius $\mathrm{R}$ with constant speed $\mathrm{v}$

Step 2: To find

We have to find the time period of the motion.

Step 3: Calculation

1. A magnetic field is a vector field that explains the magnetic influence on moving charges, currents, and magnetic materials.
2. In a magnetic field, a moving charge receives a force that is perpendicular to both its velocity and the magnetic field.

Force on a charged particle moving in a magnetic field.

$$\begin{gathered} F=q\left(v\times B\right) \\ =qvB\sin \theta \end{gathered}$$

Where, force on a charged particle is $F$, charge on the particle is $q$, the velocity of the particle is $v$ and the magnetic field is $B$.

Since the velocity of the charged particle is perpendicular to the magnetic field. Therefore, the angle between the velocity and magnetic field is ${90}^{\circ }$.

$$\begin{gathered} F=qvB\sin {90}^{\circ } \\ =qvB \end{gathered}$$

Force on a particle moving in a circular path is given by,

$F=\frac{m{v}^2}{R}$

where, the mass of the particle is $m$, and $R$ is the radius of the circular path.

Equate the value of $F$.

$\begin{array}{rcl}qvB& =& \frac{m{v}^2}{R}\\ & \Rightarrow & R=\frac{mv}{qB}\end{array}$

The velocity of the charged particle is given by,

$v=R\omega$

Where angular speed of the particle is $\omega$.

$\begin{array}{rcl}v& =& \frac{mv\omega }{qB}\\ & \Rightarrow & \omega =\frac{qB}{m}\end{array}$

The time period for the particle is given by,

$\begin{array}{rcl}T& =& \frac{2\mathrm{\pi }}{\omega }\\ & =& \frac{2\mathrm{\pi }}{{\displaystyle \raisebox{1ex}{$qB$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}}\\ & =& \frac{2\mathrm{\pi m}}{qB}\end{array}$

Therefore, the time period of the particle is independent of the $R$ and $v$.

Hence, option (C) is correct.